When 1.0 mole of acetic acid is diluted with water to a volume of 1.0 L at 25 °C, 0.42% of the acetic acid ionizes to form acetate ion and hydronium ion.
CH₃CO₂H(aq) + H₂O( $\ell$ ) $\rightleftharpoons$ CH₃CO₂^-(aq) + H₃O⁺(aq)
What percentage of the acid ionizes when 0.75 mole of acetic acid is diluted with water to 1.0 L at 25 °C?

Question

When 1.0 mole of acetic acid is diluted with water to a volume of 1.0 L at 25 °C, 0.42% of the acetic acid ionizes to form acetate ion and hydronium ion.
CH₃CO₂H(aq) + H₂O(

\ell

)

\rightleftharpoons

CH₃CO₂^-(aq) + H₃O⁺(aq)
What percentage of the acid ionizes when 0.75 mole of acetic acid is diluted with water to 1.0 L at 25 °C?

Quizplus · Accepted Answer

The Answer of When 1.0 mole of acetic acid is diluted with water to a volume of 1.0 L at 25 °C, 0.42% of the acetic acid ionizes to form acetate ion and hydronium ion. CH₃CO₂H(aq) + H₂O( $\ell$ ) $\rightleftharpoons$ CH₃CO₂^-(aq) + H₃O⁺(aq) What percentage of the acid ionizes when 0.75 mole of acetic acid is diluted with water to 1.0 L at 25 °C?

When 10 Mole of Acetic Acid Is Diluted with Water to to a Volume